If $\gamma$ is a smooth curve in $R^2$, length parametrized with respect to Euclidean metric, then the integral of $|\ddot{\gamma}|$ over the curve gives the amount the tangent vector rotates (if it is a closed immersed curve you get $2\pi k$ for some itneger $k$). Now instead of the Euclidean curvature, if you use the geodesic curvature of the curve with respect to some metric $g$ (assuming now the curve is length parametrized with respect to this metric) you get that the geodesic curvature is $\dot{\phi} + \omega(\dot{\gamma})$, where $\omega$ is the connection 1-form with respect to some orthogonal basis $E_1,E_2$ and $\dot{\gamma}=cos(\phi)E_1 + sin(\phi)E_2$. Now if you integrate this quantity over the curve, you get something which equals to amount of rotaion of the tangent vector + integral of the connection 1-form evaluated on $\dot{\gamma}$. So what does this second part of the integral geometrically mean?
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It is telling you the rate at which the moving frame $E_1,E_2$ is twisting from being parallel ("fixed") along the curve. In particular, it vanishes when $E_1$ is parallel along the curve.
answered Oct 11, 2016 at 17:22
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$\begingroup$ So I am assuming here $E_1(t),E_2(t)$ are two vector-fields defined along the curve such that one corresponds with the tangent direction and the other one corresponds to the normal direction. We also have $F_1(t),F_2(t)$ which are parallel transports of $E_1(0), E_2(0)$ along the curve. So you are saying this integral measures how much the first frame rotates with respect to the second one? $\endgroup$– SinaOct 21, 2016 at 9:59
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$\begingroup$ No, any orthonormal frame along the curve. I guess you were thinking of a constant frame, in which case the connection form vanishes. $\endgroup$ Oct 21, 2016 at 16:22
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$\begingroup$ Okay sorry so it is the frame that I wrote, i.e the one with respect to which the connection coefficients are defined. I get it now. $\endgroup$– SinaNov 11, 2016 at 18:16